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          CCPC Wannafly Winter Camp Day3
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        <p>最近比赛有点诡异，一直只有签到水准，人都没了。不禁开始思考知识面有点狭窄了，加油，继续补题。本次A题与E题不讲，补C题与G题。<a id="more"></a></p>
<h1 id="G火山哥周游世界"><a href="#G火山哥周游世界" class="headerlink" title="G火山哥周游世界"></a>G火山哥周游世界</h1><p>赛时想到树，但是没有想到是树形dp，主要是这里的根实在太多了，看了题解，原来换根$dp$完全可以解决这种题型，而且常用于解决这些问题。<br>该类题目的特点是：给定一个树形结构，需要以每个节点为根进行一系列统计。完全符合。</p>
<h2 id="解题思路"><a href="#解题思路" class="headerlink" title="解题思路"></a>解题思路</h2><p>经过$k$个点的最短路径很容易看出来就是从点$s$出发经过所有点后在一个节点上停下来，那么距离就是$dis = sum[s]-dept[s]$其中$sum[s]$表示$s$到经过$k$点后返回原点的距离，而$dept[s]$表示从$s$开始走，能到的$k$个点中最远的那个。<br>问题就分解成了两个子问题。</p>
<ol>
<li>从当前点到达k个点后返回自身距离。</li>
<li>从当前点出发能走到的最远的距离。</li>
</ol>
<p>对于这两个问题，我们先看问题1，首先单独考虑一个点,一个点时直接暴力$dfs$就好了。<br>扩展来看多个点。我们在一个点$dfs$时构建了一棵树，考虑父节点(上一棵树的根节点)与子节点的关系，对于父节点来说，他的答案是已知的，而更新到子节点我们也就只要考虑如何更新子节点，子节点维护了子节点下面的所有点的距离，根据父节点我们也就可以更新该节点往上走的所有值，同时判断是否加上一个边$(u \rightarrow v)$。<br>来看更新过程<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br></pre></td><td class="code"><pre><span class="line">if(m-sz[v]==0)&#123;//v上面没有有需要经过的点</span><br><span class="line">	dp[v]=dp[u]-ed[i].w;把边权去掉</span><br><span class="line">&#125;</span><br><span class="line">else if(sz[v]==0)&#123;//v下面有需要经过的点</span><br><span class="line">	dp[v]=dp[u]+ed[i].w;增加边权</span><br><span class="line">&#125;</span><br><span class="line">else dp[v]=dp[u];//该点是必要点(必须经过)所以边权不变。</span><br></pre></td></tr></table></figure></p>
<p>子问题2：<br>当前点能到的最远距离。<br>也是首先来看单个点，我们可以知道，对于单个点来说，直接大力$dfs$就好了。<br>接下来，我们来看多个点，我们首先已经处理一个点，考虑以他为父节点，那么父节点最远距离已知，我们将一棵按照两个节点$u \rightarrow v$割成两颗子树，分别以$u$和$v$为父节点。那么当前节点v的能到的$dept=max(v树里面能走到的最远的距离,u树能走到的最远距离+u \rightarrow v)$<br>而$v$树里面最远的距离在第一次$dfs$就可以知道，明显可以发现一个道理，不管根节点在$v$的上面怎么变，他的父节点一直不变，且$v$下面的节点结构也不变！(<del>是你爹终归是爹</del>)。<br>$u$树的最远距离有两种可能</p>
<ol>
<li>经过点$v$，那就有点尴尬了，我们考虑一条不经过$v$的次长路径和$u$往上走的最长路径更新最远距离</li>
<li>不经过点$v$，那直接就是，$u$往上走的最远距离和它往不经过v的最长路径更新最远距离。</li>
</ol>
<p>可能会问为什么是不能经过$v$的最长路径，这是$u$树(以$u$为根节点，断掉$u \rightarrow v$)边的树。<br>来看转移方程<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line">//注意，如果上面没有树，就不要去寂寞的更新了。</span><br><span class="line">if(vis[u]==v&amp;&amp;m-sz[v]!=0)&#123;</span><br><span class="line">    up[v]=max(up[u],deep1[u])+ed[i].w;</span><br><span class="line">&#125;else if(m-sz[v]!=0)&#123;</span><br><span class="line">    up[v]=max(up[u],deep0[u])+ed[i].w;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>
<p>来看全部代码<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br><span class="line">75</span><br><span class="line">76</span><br><span class="line">77</span><br><span class="line">78</span><br><span class="line">79</span><br><span class="line">80</span><br></pre></td><td class="code"><pre><span class="line">#include &lt;stdio.h&gt;</span><br><span class="line">#include &lt;algorithm&gt;</span><br><span class="line">#include &lt;string.h&gt;</span><br><span class="line">#include &lt;queue&gt;</span><br><span class="line">#include &lt;vector&gt;</span><br><span class="line">using namespace std;</span><br><span class="line">const int N = 1e6+5;</span><br><span class="line">typedef long long ll;</span><br><span class="line">#define rep(i,a,b) for(i=(a);i&lt;=b;i++)</span><br><span class="line">#define pt(a) printf(&quot;%d\n&quot;,(a))</span><br><span class="line">struct ED&#123;</span><br><span class="line">	int pre,to;</span><br><span class="line">	ll w;</span><br><span class="line">&#125;ed[N];</span><br><span class="line">int head[N],tot=0,vis[N],sz[N],n,m;</span><br><span class="line">ll deep0[N],dp[N],deep1[N],ans[N],dept[N],up[N];</span><br><span class="line">void add(int u,int v,int w)&#123;</span><br><span class="line">	ed[++tot].pre=head[u];</span><br><span class="line">	ed[tot].to=v;</span><br><span class="line">	ed[tot].w=w;</span><br><span class="line">	head[u]=tot;</span><br><span class="line">&#125;</span><br><span class="line">void dfs0(int u,int fa)&#123;</span><br><span class="line">	for(int i=head[u];i;i=ed[i].pre)&#123;</span><br><span class="line">		int v=ed[i].to;</span><br><span class="line">		if(fa==v) continue;</span><br><span class="line">		dfs0(v,u);</span><br><span class="line">		sz[u]+=sz[v];</span><br><span class="line">		if(sz[v])&#123;</span><br><span class="line">			dp[u]+=dp[v]+ed[i].w;</span><br><span class="line">			if(deep0[v]+ed[i].w&gt;deep0[u])&#123;</span><br><span class="line">				vis[u]=v;</span><br><span class="line">				deep1[u]=deep0[u];</span><br><span class="line">				deep0[u]=ed[i].w+deep0[v];//更新最长路</span><br><span class="line">			&#125;else deep1[u]=max(deep1[u],deep0[v]+ed[i].w);//保证了最长路与次长路不在同一个子节点上面</span><br><span class="line">		&#125;</span><br><span class="line">	&#125;</span><br><span class="line">&#125;</span><br><span class="line">void dfs1(int u,int fa)&#123;</span><br><span class="line">	//printf(&quot;%d %d %d %d\n&quot;,u,dp[u],deep0[u],up[u]);</span><br><span class="line">	dept[u]=max(deep0[u],up[u]);</span><br><span class="line">	ans[u]=(dp[u]*2-dept[u]);</span><br><span class="line">	for(int i=head[u];i;i=ed[i].pre)&#123;</span><br><span class="line">		int v=ed[i].to;</span><br><span class="line">		if(fa==v) continue;</span><br><span class="line">		if(m-sz[v]==0)&#123;</span><br><span class="line">			dp[v]=dp[u]-ed[i].w;</span><br><span class="line">		&#125;</span><br><span class="line">		else if(sz[v]==0)&#123;</span><br><span class="line">			dp[v]=dp[u]+ed[i].w;</span><br><span class="line">		&#125;</span><br><span class="line">		else dp[v]=dp[u];</span><br><span class="line">		if(vis[u]==v&amp;&amp;m-sz[v]!=0)&#123;</span><br><span class="line">			up[v]=max(up[u],deep1[u])+ed[i].w;</span><br><span class="line">		&#125;else if(m-sz[v]!=0)&#123;</span><br><span class="line">			up[v]=max(up[u],deep0[u])+ed[i].w;</span><br><span class="line">		&#125;</span><br><span class="line">		dfs1(v,u);</span><br><span class="line">	&#125;</span><br><span class="line">&#125;</span><br><span class="line">int main()</span><br><span class="line">&#123;</span><br><span class="line">	int i,j,k,t=0;</span><br><span class="line">	int u,v,w;</span><br><span class="line">	scanf(&quot;%d %d&quot;,&amp;n,&amp;m);</span><br><span class="line">	tot=0;</span><br><span class="line">	rep(i,1,n-1)&#123;</span><br><span class="line">		scanf(&quot;%d %d %d&quot;,&amp;u,&amp;v,&amp;w);</span><br><span class="line">		add(u,v,w);</span><br><span class="line">		add(v,u,w);</span><br><span class="line">	&#125;</span><br><span class="line">	rep(i,1,m) scanf(&quot;%d&quot;,&amp;u),sz[u]=1;</span><br><span class="line">	dfs0(1,1);</span><br><span class="line">	dfs1(1,1);</span><br><span class="line">	rep(i,1,n)&#123;</span><br><span class="line">		printf(&quot;%lld\n&quot;,ans[i]);</span><br><span class="line">	&#125;</span><br><span class="line">	//system(&quot;pause&quot;);</span><br><span class="line">	return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>
<h1 id="无向图定向"><a href="#无向图定向" class="headerlink" title="无向图定向"></a>无向图定向</h1><h2 id="迪尔沃斯定理"><a href="#迪尔沃斯定理" class="headerlink" title="迪尔沃斯定理"></a>迪尔沃斯定理</h2><p>定理指出：<strong>对于任意有限偏序集，其最长链中元素的数目必等于其最小反链划分中反链的数目</strong>。<br>具体应用有</p>
<ol>
<li>求一个序列的上升子序列的数量，我们可以求其最长下降子序列的长度就是其上升子序列的数量。<a href="https://www.luogu.com.cn/problem/P1020" target="_blank" rel="noopener">题目链接</a></li>
</ol>
<p>这一题就当是开了眼，蛮玄乎的。<br>我们暴力枚举每一个点的颜色，注意其一条边之间的两个点颜色不得相同，当颜色使用的最少的时候，就是其最小反链的数目+1，直接输出就好了。<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br></pre></td><td class="code"><pre><span class="line">#include &lt;stdio.h&gt;</span><br><span class="line">#include &lt;algorithm&gt;</span><br><span class="line">#include &lt;string.h&gt;</span><br><span class="line">#include &lt;queue&gt;</span><br><span class="line">#include &lt;vector&gt;</span><br><span class="line">using namespace std;</span><br><span class="line">const int N = 17+5;</span><br><span class="line">typedef long long ll;</span><br><span class="line">#define rep(i,a,b) for(int i=(a);i&lt;=b;i++)</span><br><span class="line">#define pt(a) printf(&quot;%d\n&quot;,(a))</span><br><span class="line">struct ED&#123;</span><br><span class="line">	int pre,to,w;</span><br><span class="line">&#125;ed[N*2];</span><br><span class="line">int head[N],tot,ans=30,n,m,siz=0,clo[N];</span><br><span class="line">void add(int u,int v)&#123;</span><br><span class="line">	ed[++tot].pre=head[u];</span><br><span class="line">	ed[tot].to=v;</span><br><span class="line">	head[u]=tot;</span><br><span class="line">&#125;</span><br><span class="line">bool check(int u,int flag)&#123;</span><br><span class="line">	for(int i=head[u];i;i=ed[i].pre)&#123;</span><br><span class="line">		if(clo[ed[i].to]==flag) return 1;</span><br><span class="line">	&#125;</span><br><span class="line">	return 0;</span><br><span class="line">&#125;</span><br><span class="line">void dfs(int u,int res)&#123;</span><br><span class="line">	if(res&gt;ans) return ;</span><br><span class="line">	if(u&gt;n)&#123;</span><br><span class="line">		ans=min(ans,res);</span><br><span class="line">		return ;</span><br><span class="line">	&#125;</span><br><span class="line">	rep(j,1,res)&#123;</span><br><span class="line">		if(check(u,j)) continue;</span><br><span class="line">		clo[u]=j;</span><br><span class="line">		dfs(u+1,res);</span><br><span class="line">	&#125;</span><br><span class="line">	clo[u]=res+1;</span><br><span class="line">	dfs(u+1,res+1);</span><br><span class="line">	clo[u]=0;</span><br><span class="line">&#125;</span><br><span class="line">int main()</span><br><span class="line">&#123;</span><br><span class="line">	int i,j,k,t=0;</span><br><span class="line">	scanf(&quot;%d %d&quot;,&amp;n,&amp;m);</span><br><span class="line">	ans=m+1;</span><br><span class="line">	siz=1,tot=0;</span><br><span class="line">	rep(i,1,m)&#123;</span><br><span class="line">		int u,v;</span><br><span class="line">		scanf(&quot;%d %d&quot;,&amp;u,&amp;v);</span><br><span class="line">		add(v,u);	</span><br><span class="line">		add(u,v);</span><br><span class="line">	&#125;</span><br><span class="line">	dfs(1,1);</span><br><span class="line">	printf(&quot;%d\n&quot;,ans-1);</span><br><span class="line">	//system(&quot;pause&quot;);</span><br><span class="line">	return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>
<h1 id="参考资料"><a href="#参考资料" class="headerlink" title="参考资料"></a>参考资料</h1><ol>
<li><a href="https://www.cnblogs.com/laoguantongxiegogo/p/12490457.html" target="_blank" rel="noopener">迪尔沃斯定理</a></li>
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